20 students are standing in a line to buy food at football game the first three students spend 24$ on food how much do thr first
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
Given that the probability that an American company will open a location in Shanghai, China is 0.7.
The probability that it will open a location in Beijing, China is 0.4.
The probability that it will open a location in either Shanghai or Beijing or both is 0.8.
the probability that the company will open a location in:
(a) both Shanghai and Beijing :
P(both) = 0.3
(using addition theory of probability for union of sets)
(b) neither Shanghai nor Beijing
= 1- P(AUB)
(c) Beijing but not in Shanghai?
P(Beiging)_P(both) =
(d) Shanghai given that it opens a location in Beijing
=P(Both)/P(Beijing) =
The angle is 127°, so the Pythagorean theorem does not apply.
Generally, you would use the Law of Cosines to solve for the third side when given two sides and the angle between them.
EF^2 = 7^2 +15^2 -2*7*15*cos(127°)
EF ≈ √(400.3812) ≈ 20.00953
My guess is that you're expected to use 20 for the length of the 3rd side.
By Heron's formula, s = (7 +15 +20)/2 = 21
Area = √(21*(21 -7)*(21 -15)*(21 -20)) = √(21*14*6*1) = √1764 = 42 . . . . yd^2
_____
More directly,
.. Area = (1/2)*7*15*sin(127°) ≈ 41.928 . . . . yd^2
Generally, you would use the Law of Cosines to solve for the third side when given two sides and the angle between them.
EF^2 = 7^2 +15^2 -2*7*15*cos(127°)
EF ≈ √(400.3812) ≈ 20.00953
My guess is that you're expected to use 20 for the length of the 3rd side.
By Heron's formula, s = (7 +15 +20)/2 = 21
Area = √(21*(21 -7)*(21 -15)*(21 -20)) = √(21*14*6*1) = √1764 = 42 . . . . yd^2
_____
More directly,
.. Area = (1/2)*7*15*sin(127°) ≈ 41.928 . . . . yd^2